\newproblem{lay:7_4_17}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.4.17}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Suppose $A$ is square and invertible. Find a Singular Value Decomposition of $A^{-1}$
}{
   % Solution
	Let $A=U\Sigma V^T$ be a Singular Value Decomposition of the matrix $A$. Since $A$ is invertible, $\Sigma$ is full rank. Since $A$ is square, $U$ and $V$ are square matrices, and 
	they are always orthogonal matrices. So, we have
	\begin{center}
		$A^{-1}=(U\Sigma V^T)^{-1}=(V^T)^{-1}\Sigma^{-1}U^{-1}=V\Sigma^{-1}U^T$
	\end{center}
	that is a Singular Value Decomposition of $A^{-1}$.
}
\useproblem{lay:7_4_17}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

